Now if a child has red hair, both parents have at least one recessive allele, so the possible combinations are (Aa Aa), (Aa aa), and (aa aa). If we assume, again, that mating decisions are not based on hair color, we can get the prevalence of each parental pair:

P(Aa Aa) = Aa**2

P(Aa aa) = 2 Aa aa

P(aa aa) = aa**2

And we can use those as the priors. The evidence is

E: the child has red hair

To get the likelihoods, we apply Mendelian genetics:

P(E | Aa Aa) = 0.25

P(E | Aa aa) = 0.5

P(E | aa aa) = 1.0

Finally, applying Bayes's Theorem, we have

P(Aa Aa | E) = P(Aa Aa) P(E | Aa Aa) / P(E)

And the answer is 0.737. With a little algebra we can show that

P(Aa Aa | E) = P(AA)

That is, the probability that neither parent has red hair is exactly the fraction of the population that is homozygous dominant.

Almost 75% of red-haired people come from parents with non-red hair, which might explain why a "red haired child" is a metaphor for a child that doesn't resemble his parents. In the expression, "beat like a red haired step child," some of the humor (for people who find child abuse funny) comes from the suggestion that the parentage of a red haired child is suspect.

But why should red hair be funnier than blue eyes or blonde hair? It turns out that we can answer this question mathematically. If the prevalence of red hair were higher, say 10%, most red haired people would have at least one red-haired parent, and that would be less funny.

In general, as the prevalence of the recessive phenotype increases, the potential for amusing insinuations of infidelity decreases; near 0 it drops off steeply, as shown in this figure:

Since I believe I am the first person to quantify this effect, I humbly submit that it should be called "Downey's inverse law of mailman jokes."

For more fun with probability, see Chapter 5 of my book, Think Stats, which you can read here, or buy here.

-----

If you don't get the title of this post, it is a play on "Somebody bet on the bay," a lyric from the minstrel song "Camptown Races." A bay is a horse with a reddish-brown coat, so I thought it was a pretty good fit.

3 comments:

I found it a little easier to analyze this situation by realizing that you can calculate the probability of each parent's readheadedness separately. The prior probabilities relating to the two parents are independent, and the evidence "factors" into a statement just about the mother (mother passed on "a" to child) and a statement just about the father (father passed on "a" to child). So the posterior probability for the various genetic makeups of the mother and father are independent, and you can calculate them one at a time.

Consider just the mother. The prior probabilities for her genetic makeup are what you said:

p(AA)=q^2P(Aa)=2pqP(aa)=p^2

The evidence E is that the mother passed on a to her child.

P(E | AA) = 0P(E | Aa) = 1/2P(E | aa) = 1

Turning the Bayes's theorem crank, we get

P(AA | E) = 0 (of course)P(Aa | E) = qP(aa | E) = p

So the probability that the mother didn't have red hair is q. The same reasoning works for the father, so the probability that neither had red hair is q^2.